Question

Help yet again with Trig! ( i tried it, but stuck at the end): find all possible degree solutions
2 sin ^2 x - 2 sin x -1=0

Answer 1

i tried quadratic formula

Answer 2

ok...what do we have here....

Answer 3

quads fine.... what did you get for "u"?

2u^2 -2u -1 = 0 where u = sin(x)

Answer 4

2 +- 2sqrt(2)
------------ right?
4

Answer 5

i probably messes that up in my head lol

Answer 6

4 - (4)(2)(-1)

12 = 4*3

2sqrt(3) is better

Answer 7

(1/2) +- sqrt(3)/2

right?

Answer 8

yes

Answer 9

sin(x) = (1/2) + sqrt(3)/2

sin(x) = sin(30) + cos(30) it looks like

Answer 10

and sin(x) = sin(30) - sin(60)

Answer 11

sin...cos....keep em straight!! lol

Answer 12

You need all degree solutions, yeah, not just over a fixed interval?

Answer 13

sin(x) = 2cos(30) but i might have done it in error

Answer 14

In general, \[\theta = n.180^o+(-1)^n \alpha^o\]where \[\alpha^o\]is a seed angle and\[n \in \mathbb{Z}\]You have the seed from the solution to the quadratic, \[\alpha^o = \sin^{-1}\left( \frac{1-\sqrt{3}}{2} \right)\]You have to reject the other solution since arcsine is not defined for an argument outside magnitude 1. So your solution should be,\[\theta =180^o+(-1)^n \sin^{-1}\left( \frac{1-\sqrt{3}}{2} \right)^o\] for n an integer.

Answer 15

338.51 degrees

and

201.47 degrees are the 2 angles that would produce the proper sin(x) value.... right? then we add k2pi periods to those toget all the values....maybe :)

Answer 16

Sorry, in my last result, there's a typo. It should read:\[\theta = n.180^o+(-1)^n \sin ^ {-1} \left( \frac{1-\sqrt{3}}{2} \right)\]

Answer 17

Yeah, amstre, you're on the right track.

Answer 18

sorry, about late response, but thank you for helping me